Question

Show that ${\mathrm{n}}^2-1$is divisible by $8$, if $\mathrm{n}$ is an odd positive integer.

Answer 1

Explanation:

Given that:$\mathrm{n}$ is an odd positive integer

To Prove:

${\mathrm{n}}^2-1$ is divisible by $8$ if $\mathrm{n}$ is an odd integer.

We know that,

Odd number is in the form of$(4\mathrm{q}+1)$ where $\mathrm{q}$ is a natural number,

When $\mathrm{n}=$$(4\mathrm{q}+1)$

so, $\mathrm{n}²-1=(4\mathrm{q}+1)²-1$

$\mathrm{n}²-1=16\mathrm{q}²+8\mathrm{q}+1-1$ $\left[{\because (\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\right]$

$=16\mathrm{q}²+8\mathrm{q}$

$=8(2\mathrm{q}+1)$ is divisible by $8$

When $\mathrm{n}=4\mathrm{q}+3$

Then ${\mathrm{n}}^2–1={(4\mathrm{q}+3)}^2–1$

$=16{\mathrm{q}}^2+9+24\mathrm{q}–1$ $\left[{\because (\mathrm{a}+\mathrm{b})}^2={\mathrm{a}}^2+{\mathrm{b}}^2+2\mathrm{ab}\right]$

$=16{\mathrm{q}}^2+24\mathrm{q}+8$

$=8(2{\mathrm{q}}^2+3\mathrm{q}+1)$ is divisible by $8$

It is concluded that odd number $\mathrm{n}$ is divisible by $8$ for all natural numbers.

$\mathrm{n}²-1$ is divisible by $8$ for all odd values of $\mathrm{n}$.

Hence Proved.